How do you factor #10a^2 - 14a - 12#?

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Answer 1 · 2015-06-06T13:12:49

#10a^2-14a-12=2(5a^2-7a-6)#

Use AC Method to find factors of #5a^2-7a-6# ...

#A=5#, #B=7#, #C=6#

Look for a pair of factors of #AC=5*6=30# whose difference is #B=7#.

The pair #B1=10#, #B2=3# works.

Then for each of the pairs #(A, B1)#, #(A, B2)# divide by the HCF to get the coefficients of a factor, choosing signs as appropriate:

#(A, B1) = (5, 10) -> (1, 2) -> (a-2)#
#(A, B2) = (5, 3) -> (5, 3) -> (5a+3)#

So:

#10a^2-14a-12=2(5a^2-7a-6) = 2(a-2)(5a+3)#